What is the solution for 4y^{''}+4y^'+9y=0 ?

The solution for 4y^{''}+4y^'+9y=0 is y=e^{-t/2}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4y^{''}+4y^'+9y=0

4 y^{^{''}} + 4 y^{^{'}} + 9 y = 0

y = e^{- \frac{t}{2}} (c_{1} cos (2 t) + c_{2} sin (2 t))

Solution steps

4 y^{''} (t) + 4 y^{'} (t) + 9 y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (2 t) + c_{2} sin (2 t))

y = e^{- \frac{t}{2}} (c_{1} cos (2 t) + c_{2} sin (2 t))

\int_{2}^{3} \frac{1}{x ^{2} - 1} d x

\int (y - 2)^{2} d y

\frac{d}{d x} (\frac{( x ^{2} - 5 ) ^{3}}{x ^{2} + 4})

\int + 4 e^{- 7 x} d x

\frac{\partial}{\partial x} (3 z e^{x y})

What is the solution for 4y^{''}+4y^'+9y=0 ?
The solution for 4y^{''}+4y^'+9y=0 is y=e^{-t/2}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))